Chapter 4: Q. 130 (page 532)

Find an equation of a line parallel to the line $y = \frac{1}{2} x - 3$ that contains the point $(6, 4)$ .

Short Answer

Expert verified

The equation of the line is $y = \frac{1}{2} x + 1$ .

Step by step solution

Step 1. Given Information.

Equation of the line, $y = \frac{1}{2} x - 3$ .

Point is $(6, 4)$ .

Step 2. Finding the slope.

The slope-intercept form is,

$y = m x + b$

Comparing the equation $y = \frac{1}{2} x - 3$ we get,

$m_{1} = \frac{1}{2}$

The slope of the parallel lines is equal.

$m_{1} = m_{2} \Rightarrow m_{2} = \frac{1}{2}$

Step 3. Equation of the line.

The point-slope form is,

$(y - y_{1}) = m (x - x_{1})$

Inputting the value,

$(y - 4) = \frac{1}{2} (x - 6) y - 4 = \frac{1}{2} x - 3 y = \frac{1}{2} x + 1$

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Find an equation of a line parallel to the line $y = \frac{1}{2} x - 3$ that contains the point $(6, 4)$ .

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Finding the slope.

Step 3. Equation of the line.

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Find an equation of a line parallel to the line y=12x-3that contains the point (6,4).

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Finding the slope.

Step 3. Equation of the line.

One App. One Place for Learning.

Most popular questions from this chapter

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Find an equation of a line parallel to the line $y = \frac{1}{2} x - 3$ that contains the point $(6, 4)$ .